Derivative of Normalization Map has null space of dimension 1

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Derivative of Normalization Map has null space of dimension 1

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myth17

#1 h Mar 20, 2025, 5:16 PM

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Let $f(\vec{x}) = \frac{\vec{x}}{||\vec{x}||}$ be defined on $\mathbb{R}^n \setminus \{\vec{0}\}$ . Show that the dimension of the kernel of $Df_{\vec{x}}$ for any $\vec{x} \in \mathbb{R}^n \setminus \{\vec{0}\}$ is $1$ .

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#2 h Mar 20, 2025, 8:16 PM • 1 Y

Y by myth17

The difficult part is the calculation of $Df$ : write $\beta(x)=\sum_1^n x_k^2$ and $\alpha=\sqrt{\beta}$ then $f(x)={1\over \alpha}x$ . Note that $\nabla\beta(x)=2x$ . Hence $Df={-1\over {\alpha^2}}\cdot {1\over {2\alpha}}2xx^T+{1\over \alpha}I={{\alpha^2 I-xx^T}\over {\alpha^3}}$ .
For the kernel of $Df$ we can forget about the factor ${1\over {\alpha^3}}$ .
Hence $Df(v)=0\iff xx^T v=\alpha^2 v\iff (x,v)x=\alpha^2 v \implies v\in L(x)$ .
And for $v=tx$ we have $(x,v)x=(x,tx)x=t||x||^2 x=t\alpha^2 x=\alpha^2 v$ . Hence $\ker(Df)=L(x)$ .
It was a priori clear that $x\in \ker(Df)$ since $f$ is constant on the halfline from 0 to $x$ .

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myth17

#3 h Mar 21, 2025, 2:45 AM

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alexheinis wrote:

The difficult part is the calculation of $Df$ : write $\beta(x)=\sum_1^n x_k^2$ and $\alpha=\sqrt{\beta}$ then $f(x)={1\over \alpha}x$ . Note that $\nabla\beta(x)=2x$ . Hence $Df={-1\over {\alpha^2}}\cdot {1\over {2\alpha}}2xx^T+{1\over \alpha}I={{\alpha^2 I-xx^T}\over {\alpha^3}}$ .
For the kernel of $Df$ we can forget about the factor ${1\over {\alpha^3}}$ .
Hence $Df(v)=0\iff xx^T v=\alpha^2 v\iff (x,v)x=\alpha^2 v \implies v\in L(x)$ .
And for $v=tx$ we have $(x,v)x=(x,tx)x=t||x||^2 x=t\alpha^2 x=\alpha^2 v$ . Hence $\ker(Df)=L(x)$ .
It was a priori clear that $x\in \ker(Df)$ since $f$ is constant on the halfline from 0 to $x$ .

Thank you very much for the response. I will write your solution here in a way that makes the steps clear to me, for my own future reference.

Observe that $2\vec{x} = \nabla(||\vec{x}||^2) = 2||\vec{x}|| \nabla ||\vec{x}|| \implies \nabla ||\vec{x}|| = \frac{\vec{x}}{||\vec{x}||}$ , so by the product rule,

$Df(\vec{x}) = D\left(\vec{x} \frac{1}{||\vec{x}||}\right) = \vec{x}D\left(\frac{1}{||\vec{x}||}\right) + \frac{D\vec{x}}{||\vec{x}||} = \vec{x}\left(\nabla\left(||\vec{x}||^{-1}\right)\right)^T + \frac{I}{||\vec{x}||} = -\vec{x}||\vec{x}||^{-2}\left(\nabla\left(||\vec{x}||\right)\right)^T + \frac{I}{||\vec{x}||}$
$= -\vec{x}||\vec{x}||^{-2}\left(\frac{\vec{x}}{||\vec{x}||}\right)^T + \frac{I}{||\vec{x}||} = \frac{-\vec{x}\vec{x}^T + ||\vec{x}||^2}{||\vec{x}||^3}.$
Then,

$\vec{v} \in \text{ker}(Df(\vec{x})) \iff (-\vec{x}\vec{x}^T + ||\vec{x}||^2)\vec{v} = 0 \iff \vec{x}(\vec{x} \cdot \vec{v}) = ||\vec{x}||^2\vec{v} \iff \vec{v} = \frac{\vec{x} \cdot \vec{v}}{||\vec{x}||^2}\vec{x} = \text{proj}_{\vec{x}}\vec{v} \iff \vec{v} \in \mathbb{R}\vec{x},$
implying $\text{ker}(Df(\vec{x})) = \mathbb{R}\vec{x}$ . Again, much appreciate your response.

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#4 h Mar 21, 2025, 10:26 AM • 2 Y

Y by alexheinis, myth17

$f:x\in\mathbb{R}^n \setminus \{\vec{0}\}\mapsto S_{n-1}\subset \mathbb{R}^n$ .
Then $im(Df_x)\subset T_{f(x)}(S_{n-1})$ , the tangent space to $S_{n-1}$ in the point $f(x)$ .
On the other hand, $f_{|{S_{n-1}}}=id$ . Thus $im(Df_x))=T_{f(x)}(S_{n-1})$ and $rank(Df_x)=n-1$ .
Thus $dim(\ker(Df_x))=1$ ; more precisely, since $f(x+\lambda x)=f(x)$ , we deduce that $\ker(Df_x)=span[x]$ .

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myth17

#5 h Mar 21, 2025, 1:43 PM

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loup blanc wrote:

$f:x\in\mathbb{R}^n \setminus \{\vec{0}\}\mapsto S_{n-1}\subset \mathbb{R}^n$ .
Then $im(Df_x)\subset T_{f(x)}(S_{n-1})$ , the tangent space to $S_{n-1}$ in the point $f(x)$ .
On the other hand, $f_{|{S_{n-1}}}=id$ . Thus $im(Df_x))=T_{f(x)}(S_{n-1})$ and $rank(Df_x)=n-1$ .
Thus $dim(\ker(Df_x))=1$ ; more precisely, since $f(x+\lambda x)=f(x)$ , we deduce that $\ker(Df_x)=span[x]$ .

Thank you very much for your response. I will write your solution in a way that is clear to me for my own reference.

Observe that $f(\vec{x}) = \frac{\vec{x}}{||\vec{x}||}: \mathbb{R}^n \setminus \{\vec{0}\} \to S_{n - 1}$ , so its derivative is a map between the tangent spaces of $\vec{x}$ and $f(\vec{x})$ , implying $Df_{\vec{x}}(T_{\vec{x}}(\mathbb{R}^n \setminus \{\vec{0}\})) \subseteq T_{f(\vec{x})}S_{n - 1}$ . We can see the two sets are equal as follows. For any $r > 0$ and $||\vec{x}|| = r$ , consider $f(\vec{x}) = \vec{x}/r$ as a map from $rS_{n - 1} \to S_{n - 1}$ . Differentiating gives $Df_{\vec{x}} = \frac{I}{r}: T_{\vec{x}}(rS_{n - 1}) \to T_{f(\vec{x})}S_{n - 1}$ . Both tangent spaces are isomorphic to $\mathbb{R}^{n - 1}$ because $rS_{n - 1}$ and $S_{n - 1}$ are $(n - 1)$ -dimensional, and since $\frac{I}{r}: \mathbb{R}^{n - 1} \to \mathbb{R}^{n - 1}$ is surjective, we have $Df_{\vec{x}}(T_{\vec{x}}(rS_{n - 1})) = T_{f(\vec{x})}S_{n - 1}$ . The left side is in $Df_{\vec{x}}(T_{\vec{x}}(\mathbb{R}^n \setminus \{\vec{0}\}))$ so we must have $Df_{\vec{x}}(T_{\vec{x}}(\mathbb{R}^n \setminus \{\vec{0}\})) = T_{f(\vec{x})}S_{n - 1}$ as well. Equivalently, $\text{rank}(Df_{\vec{x}}) = n - 1$ . Since $f(\lambda \vec{x}) = f(\vec{x})$ for any $\lambda \neq 0$ , we know $f$ is constant along $\mathbb{R}_{\neq 0}\vec{x}$ so zero is the result when its derivative is multiplied by any vector on that line. This implies $\mathbb{R}\vec{x} \subseteq \text{ker}(Df_{\vec{x}})$ . By the rank-nullity theorem, $\text{ker}(Df_{\vec{x}})$ has dimension $n - \text{dim}(\text{rank}(Df_{\vec{x}})) = 1$ , so we must have equality.

Again, much appreciate your response.

This post has been edited 1 time. Last edited by myth17, Mar 21, 2025, 1:50 PM
Reason: Reasoning in last step.

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